<u>Answer:</u> The molar concentration of ammonia is 3.008 M
<u>Explanation:</u>
To calculate the concentration of base, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is
are the n-factor, molarity and volume of base which is
We are given:
Putting values in above equation, we get:
Hence, the molar concentration of ammonia is 3.008 M
Answer:
4d orbital.
Explanation:
Hello!
In this case, since zirconium's atomic number is 40, we fill in the electron configuration up to 40 as shown below:
Thus, the orbital 4d is partially filled.
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This should help :)
Example 1: A 36.0 g sample of water is initially at 10.0 °C.
How much energy is required to turn it into steam at 200.0 °C? (This
example starts with a temperature change, then a phase change followed
by another temperature change.)
Solution:
<span>q = (36.0 g) (90.0 °C) (4.184 J g¯1 °C¯1) = 13,556 J = 13.556 kJ
q = (40.7 kJ/mol) (36.0 g / 18.0 g/mol) = 81.4 kJ
q = (36.0 g) (100.0 °C) (2.02 J g¯1 °C¯1) = 7272 J = 7.272 kJ
q = 102 kJ (rounded to the appropriate number of significant figures)
</span>
To do this, we need molar mass, as well as Avogadro's Number.
To start, 1.60 x 10-12 grams C10H16O. Molar mass = 152.23 g/mol
Divide 1.60 x 10-12 grams by 152.23 g/mol to get a number in moles.
You should get 1.05 x 10-14 moles
From here, we use Avogadro's Numner, 6.022 x 10^23 molecules/mole.
Multiply our moles, 1.05 x 10-14 by 6.022 x 10^23
Answer:
6.33 x 10^9 molecules.
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Answer:
do you speak english? just curious?
Explanation: